Divide using synthetic division $$ ( 3x^{3}-2x^{2}+5 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&3&-2&0&5\\& & -9& 33& \color{black}{-99} \\ \hline &\color{blue}{3}&\color{blue}{-11}&\color{blue}{33}&\color{orangered}{-94} \end{array} $$The solution is:
$$ \frac{ 3x^{3}-2x^{2}+5 }{ x+3 } = \color{blue}{3x^{2}-11x+33} \color{red}{~-~} \frac{ \color{red}{ 94 } }{ x+3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&3&-2&0&5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 3 }&-2&0&5\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 3 } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&3&-2&0&5\\& & \color{blue}{-9} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -11 } $
$$ \begin{array}{c|rrrr}-3&3&\color{orangered}{ -2 }&0&5\\& & \color{orangered}{-9} & & \\ \hline &3&\color{orangered}{-11}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -11 \right) } = \color{blue}{ 33 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&3&-2&0&5\\& & -9& \color{blue}{33} & \\ \hline &3&\color{blue}{-11}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 33 } = \color{orangered}{ 33 } $
$$ \begin{array}{c|rrrr}-3&3&-2&\color{orangered}{ 0 }&5\\& & -9& \color{orangered}{33} & \\ \hline &3&-11&\color{orangered}{33}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 33 } = \color{blue}{ -99 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&3&-2&0&5\\& & -9& 33& \color{blue}{-99} \\ \hline &3&-11&\color{blue}{33}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -99 \right) } = \color{orangered}{ -94 } $
$$ \begin{array}{c|rrrr}-3&3&-2&0&\color{orangered}{ 5 }\\& & -9& 33& \color{orangered}{-99} \\ \hline &\color{blue}{3}&\color{blue}{-11}&\color{blue}{33}&\color{orangered}{-94} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}-11x+33 } $ with a remainder of $ \color{red}{ -94 } $.